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A particle moves in xy plane with constant acceleration a directed along negative y-axis.The equation of path of the particle has form y=bx-cx^2,where a and b are positive constants.Find the velocity of the particle at the origin of coordinates. ANS:under root[a(1+b^2)/2c] REQUEST:Please...submit with sufficient explanation.

A particle moves in xy plane with constant acceleration a directed along negative y-axis.The equation of path of the particle has form y=bx-cx^2,where a and b are positive constants.Find the velocity of the particle at the origin of coordinates.
ANS:under root[a(1+b^2)/2c]
REQUEST:Please...submit with sufficient explanation.

Grade:12th Pass

2 Answers

Akash Kumar Dutta
98 Points
8 years ago

Dear Sammer,
Comparing the eq. with projectile from earth=>y=x.tan m -x^2tan m/R...where R=Range
We get a=g...b=tan m.....c=tan m/R
hence c=b/R=>R=b/c
But R=u^2sin 2m / a
Hence u^2=ab/2c(sin m).(cos m)
as tan m=b,putting the value of sin m and cos m
the eq becomes
u^2=a(1+b^2)/2c
Hence u=root[a(1+b^2)/2c]
ANS.
Regards.

Tushar
15 Points
2 years ago
Y=bX-cX^2
Differentiating it
 (     V(y)=bV(x)-2cXV(x).    )-#1
At origin X=0
So.(    V(y)=bV(x).   )-#2
Differentiating #1 further
a(y)=ba(x)-2c(V(x))^2-2ca(y)X
At origin X=0 and acc. Is along -ve Y so a(x) =0
-a(y)=a(total)
a=2c(V(x))^2
So V(x)^2= a/2c
From #2.  V(y)^2=b^2.a/2c
Total V = (.  V(x)^2+V(y)^2. )^1/2
So V= (  a(p^2+1)/2q.  )1/2

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